How to use these questions
Work through each question in order and commit to one option before reading its explanation. For every wrong answer, identify the step that failed: drawing the current path, choosing a sign, applying a formula, or checking units. Getting the numerical answer right by a lucky guess is not the same as being able to reproduce the method.
These questions cover common circuit-analysis skills rather than the syllabus of a particular examination. Check your course materials for the topics and conventions you are expected to know. The first MySummaries practice item tests a series network.
A 12 V ideal source is connected in series with 2 kΩ and 4 kΩ resistors. What is the current through the 4 kΩ resistor?
The series resistance is 2 kΩ + 4 kΩ = 6 kΩ, so I = V/R = 12/6000 = 0.002 A = 2 mA. The strongest distractor is 3 mA, which comes from incorrectly using only the 4 kΩ resistor and ignoring the other series resistor.
In a series path, the current is the same through every component; the voltage divides between them. Check that distinction before moving on to the parallel circuit.
Questions
This item tests whether you can combine parallel branch currents while keeping track of the voltage shared by the branches.
A 12 V ideal source is connected across 6 Ω and 3 Ω resistors in parallel. What current does the source supply?
Each branch has 12 V across it. The 6 Ω branch draws 12/6 = 2 A and the 3 Ω branch draws 12/3 = 4 A, so the source supplies 6 A. The strongest distractor is 4 A: that is the current in the 3 Ω branch alone, not the total source current.
For parallel branches, add currents at the source. As a check, the equivalent resistance is 2 Ω, so 12 V across the network also gives 6 A.
The next question checks Kirchhoff’s current law at a junction, including the direction of the unknown current.
At a node, currents of 2.4 mA and 1.1 mA enter. Currents of 0.8 mA and x leave. What is x?
Kirchhoff’s current law says total current entering equals total current leaving. Thus 2.4 + 1.1 = 0.8 + x, giving x = 2.7 mA. The strongest distractor is 3.5 mA, which adds the entering currents but fails to subtract the known 0.8 mA leaving the node.
Write the entering and leaving totals before rearranging. This simple sign convention prevents many avoidable errors in larger nodal-analysis problems.
This item tests the voltage-divider relationship and whether you have identified which resistor is across the output.
A 9 V ideal source feeds a 2 kΩ resistor in series with a 4 kΩ resistor to ground. The output is measured across the 4 kΩ resistor. What is the output voltage?
The series current is 9 V/(2 kΩ + 4 kΩ) = 1.5 mA. The output across 4 kΩ is 1.5 mA × 4 kΩ = 6 V. The strongest distractor is 3 V, which assigns the larger share of the supply to the smaller resistor; voltage divides in proportion to resistance.
A quick check is that the output across the 4 kΩ resistor must be greater than half the supply, because it is the larger of the two series resistors.
The fifth question checks electrical power and the conversion between current, resistance and watts.
A 10 Ω resistor carries 2 A. How much power is dissipated in the resistor?
Use P = I²R: (2 A)² × 10 Ω = 40 W. The strongest distractor is 20 W, obtained from multiplying current by resistance without using a valid power relation. The equivalent check is V = IR = 20 V, then P = VI = 40 W.
Remember the three useful forms: P = VI, P = I²R and P = V²/R. Choose the form that uses the values given, and check that the result has units of watts.
This item tests the time constant and the voltage reached during capacitor charging from an initially uncharged state.
An initially uncharged 100 µF capacitor charges from a 5 V ideal source through a 10 kΩ resistor. What is its voltage after 1.0 s?
The time constant is τ = RC = 10,000 × 100 × 10⁻⁶ = 1.0 s. For charging, V_C = V_S(1 − e^(−t/τ)); at one time constant this is 5(1 − e⁻¹) ≈ 3.16 V. The strongest distractor is 1.84 V, which is the remaining difference from the supply, 5e⁻¹, rather than the capacitor voltage.
Distinguish the capacitor’s rising voltage from the decaying voltage difference between the supply and capacitor. At one time constant, the capacitor has reached about 63.2% of its final voltage.
The seventh question tests the magnitude of series impedance in a resistor–inductor AC circuit.
A series circuit has a 30 Ω resistor and an inductor with reactance 40 Ω at the supply frequency. The supply is 100 V RMS. What is the RMS current?
For series R–L impedance, |Z| = √(R² + X_L²) = √(30² + 40²) = 50 Ω. Therefore I = 100/50 = 2.00 A RMS. The strongest distractor is 3.33 A, found by dividing by the resistance alone and ignoring the inductive reactance.
The current lags the supply voltage in an inductive circuit. Here the phase angle is arctan(40/30), about 53.1°, which is a useful consistency check even though only the current magnitude was requested.
The final item tests source polarity in a single-loop Kirchhoff voltage law calculation.
A single loop contains a 12 V source and a 5 V source connected in opposition, with a 1.0 kΩ resistor. What is the loop current, and in which direction does it flow?
Opposing sources give a net emf of 12 − 5 = 7 V. The loop current is 7 V/1.0 kΩ = 7 mA, in the direction driven by the stronger 12 V source. The strongest distractor is 17 mA, which adds the source voltages as though they were aiding rather than opposing.
When a loop contains multiple sources, mark their polarities on a sketch before applying Kirchhoff’s voltage law. The sign of each source depends on the direction you travel around the loop.
Use your errors to choose the next topic
A practice set becomes more useful when it shows which method needs attention. The example below groups errors by circuit topic; the counts represent practice appearances, not an official examination frequency.
A topic review can be laid out like this:
For each weak area, redo one problem slowly and explain each step aloud or in writing. Then attempt a fresh problem without looking at the worked method. If you repeatedly lose the same point, turn that specific misunderstanding into a short recall prompt.
A repeated error can become a targeted reminder rather than another missed mark.
You lost this mark twice: for an initially uncharged capacitor charging through R, what fraction of its final voltage has it reached after one time constant RC?
How MySummaries helps
MySummaries can build a revision board from your circuit notes, then generate practice questions and flashcards from that material. Use the marked results to identify weak topics and revisit the relevant notes. Open MySummaries.